In a quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect each other at point $O$ such that $\frac{AO}{BO}=\frac{CO}{DO}$. Prove that $ABCD$ is a trapezium.

Step-by-step solution

1. Rearrange the given relation $\frac{AO}{BO}=\frac{CO}{DO}$ as $\frac{AO}{CO}=\frac{BO}{DO}$.
2. Construct a line segment $OE$ through $O$ parallel to $AB$, meeting side $AD$ at point $E$.
3. In $△ABD$, since $EO\parallel AB$, apply the Basic Proportionality Theorem to get $\frac{AE}{ED}=\frac{BO}{DO}$.
4. Substitute $\frac{BO}{DO}=\frac{AO}{CO}$ into the ratio equation from step 3 to obtain $\frac{AE}{ED}=\frac{AO}{CO}$.
5. In $△ADC$, since $\frac{AE}{ED}=\frac{AO}{CO}$, apply the converse of the Basic Proportionality Theorem to conclude that $EO\parallel DC$.
6. Since $EO\parallel AB$ and $EO\parallel DC$, it follows that $AB\parallel DC$.
7. Since quadrilateral $ABCD$ has one pair of parallel opposite sides ($AB\parallel DC$), $ABCD$ is a trapezium.

Concept tested: Converse of Basic Proportionality Theorem

Published 20 August 2026 · How this solution was prepared and checked

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